import pandas as pd
import matplotlib.pyplot as plt
import bambi as bmb
import seaborn as sns
from pygam.datasets import mcycle
from pygam import LinearGAM, s, f, GAM, l, utils
import numpy as np
import arviz as az
import pymc as pm


random_seed = 100

import warnings

warnings.simplefilter('ignore')

GAMs and GPs: Flexibility and Calibration

Flexible Spline models risk overfit and need to be carefully calibrated against real data.

In this blog post we’re going to dive into modelling of non-linear functions and explore some of the tooling available in the python eco-system. We’ll start by looking into Generalised Additive Models with splines in pyGAM before preceding to look Bayesian versions of spline modelling comparing the splines to Gaussian processes in Bambi and PyMC. Before showing how hierarchical bayesian models avoid some of the issues of overfit in simpler spline models.

Our interest in these models stems from their flexibility to approximate functions of arbitrary complexity. We’ll see how the methods work in the case of relatively straightforward toy example and then we’ll apply each of the methods to deriving insights into the functional form of insurance loss curves. In this application we adapt a data set discussed in Mick Cooney’s Stan case study to demonstrate the power of hierarchical spline models. Throughout we’ll draw on the discussion of these methods in Osvaldo Martin’s “Bayesian Analysis with Python” for practical details implementing these models.

All of these methods need to be assessed with respect to their in-sample model fit and their out of sample performance. How can we best calibrate the model fits to perform reasonably well out of sample?

Generalised Additive models

The canonical reference for GAMs is Simon Wood’s “Generalised Additive Models: An Introduction with R” which outlines in some detail the theoretical background of splines and univariate smoothers. The book stresses the trade-offs between the flexibility of splines and the need for cross-validation and penalised estimation methods for spline based modelling.

In R these penalised models fits can be achieved in mgcv which incorporates a wilkinson like formula syntax for model specification: y ~ s(x) + s(x1). The closest implementation in python is available in PyGam and we will adopt this package to illustrate the application of smoothing terms and the penalities.

PyGAM and Penalised Fits

In the next code block we load in an example data set on which to demonstrate univariate smoothing patterns using penalised splines. These models can be optimised by fitting differing strength penalities over a varying number of splines

X, y = mcycle(return_X_y=True)

fig, ax = plt.subplots(figsize=(10, 6))
ax.scatter(X, y)
ax.set_ylabel("Acceleration")
ax.set_xlabel("Time Step")
ax.set_title("Crash Test Dummy Acceleration \n Simulated Motorcycle Crash", fontsize=20)

Text(0.5, 1.0, 'Crash Test Dummy Acceleration \n Simulated Motorcycle Crash')

Next we fit a number of different models to account for the herky-jerky of the data generating processs. We vary the parameterisations to see how the numbers of splines and strength of the penalty help account for the variation in \(y\) over the support of \(X\).

gam1 = LinearGAM(s(0, n_splines=5)).fit(X, y)
gam2 = LinearGAM(s(0, n_splines=7)).fit(X, y)
gam3 = LinearGAM(s(0, n_splines=10)).fit(X, y)
gam4 = LinearGAM(s(0, n_splines=15)).fit(X, y)
gam5 = LinearGAM(s(0, lam=.1)).fit(X, y)
gam6 = LinearGAM(s(0, lam=.5)).fit(X, y)
gam7 = LinearGAM(s(0, lam=5)).fit(X, y)
gam8 = LinearGAM(s(0, lam=15)).fit(X, y)


def plot_fit(gam, X, y, ax, t, c1='b', c2='r'):
    XX = gam.generate_X_grid(term=0, n=500)

    ax.plot(XX, gam.predict(XX), color=c2, linestyle='--')
    ax.plot(XX, gam.prediction_intervals(XX, width=.95), color=c1, ls='--')

    ax.scatter(X, y, facecolor='gray', edgecolors='none')
    ax.set_title(f"""95% prediction interval with {t} \n LL: {gam.statistics_['loglikelihood']}""");

fig, axs = plt.subplots(4,2, figsize=(10, 20))
axs = axs.flatten()
titles = ['5_splines', '7_splines', '10_splines', '15_splines',
'lam=.1', 'lam=.5', 'lam=5', 'lam=15']
gs = [gam1, gam2, gam3, gam4, gam5, gam6, gam7, gam8]
for ax, g, t in zip(axs, gs, titles):
    plot_fit(g, X, y, ax, t)

Here we’ve seen the PyGAM package applied to fitting our model to the data. In the formula specification we see y ~ s(i) where i denotes the index of the column variable in the X data.

Over the range of of the x-axis we can see how the conditional expectation is more or less well fit to the data depending on how the penalities and complexity of the model is specified.

Optimising The Parameter Setting

We can see from the model summary what is going on under the hood. For a given model specification the summary will report a number of model-fit statistics such as the log-likelihood and the AIC.

## Naive Model manually specified splines
gam_raw = LinearGAM(s(0)).fit(X, y)
print("log_likelihood:", gam_raw.statistics_['loglikelihood'])
print("AIC:", gam_raw.statistics_['AIC'])

log_likelihood: -954.5558010528869
AIC: 1931.5965064139743

The question then becomes what changes are induced in the model as we seek to optimise these model fit statistics.

## model optimised
gam = LinearGAM(s(0),  fit_intercept=False)
gam.gridsearch(X, y)
gam.summary()

  0% (0 of 11) |                         | Elapsed Time: 0:00:00 ETA:  --:--:--

  9% (1 of 11) |##                       | Elapsed Time: 0:00:00 ETA:   0:00:00

100% (11 of 11) |########################| Elapsed Time: 0:00:00 Time:  0:00:00

LinearGAM                                                                                                 
=============================================== ==========================================================
Distribution:                        NormalDist Effective DoF:                                     11.8135
Link Function:                     IdentityLink Log Likelihood:                                  -952.2409
Number of Samples:                          133 AIC:                                             1930.1088
                                                AICc:                                            1933.0789
                                                GCV:                                              609.3811
                                                Scale:                                            512.7965
                                                Pseudo R-Squared:                                   0.7984
==========================================================================================================
Feature Function                  Lambda               Rank         EDoF         P > x        Sig. Code   
================================= ==================== ============ ============ ============ ============
s(0)                              [0.2512]             20           11.8         1.11e-16     ***         
==========================================================================================================
Significance codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

WARNING: Fitting splines and a linear function to a feature introduces a model identifiability problem
         which can cause p-values to appear significant when they are not.

WARNING: p-values calculated in this manner behave correctly for un-penalized models or models with
         known smoothing parameters, but when smoothing parameters have been estimated, the p-values
         are typically lower than they should be, meaning that the tests reject the null too readily.

Fortunately, this routine can be performed directly and results in the following differences between the naive and optimised model.

Plot the GAM fits

fig, ax = plt.subplots(figsize=(10, 6))
plot_fit(gam_raw, X, y, ax, "Unoptimised Fit", c1='orange', c2='green')
plot_fit(gam, X, y, ax, "Optimised Fit")

This is all well and good. We’ve seen an approach to modelling that can capture eccentric patterns in raw data. But how does it work and why should we care?

If you’re familiar with the nomenclature of machine learning, you should think of spline modelling as a variety of feature creation. It generates “synthetic” features over the range of the observed variable. These synthetic features are the splines in question.

Digression on the usage of “Splines”

The history of the term “spline” is related to the history of draftmanship. Historically splines were thin strips of flexible wood or plastic that could be bent or shaped around a weight or “knot” points to express a traceable curve over the space of a “numberline”. The elastic nature of the spline material allowed it to be bent around the knot points of curvature expressing a smooth or continuous bend.

The mathematical technique apes these properties by defining a curve over in an analogous way. We specify “knots” to carve up the support of the random variable \(X\) into portions that require different weighting schemes to represent the outcome \(y\)

spline

Extracting the Splines

We can extract the spline features used in the PyGAM by invoking the following commands to first identify the knot points and create the b-spline basis appropriate for the variable \(X\).

knot_edges=utils.gen_edge_knots(X,dtype='numerical')
knots=np.linspace(knot_edges[0],knot_edges[-1],len(gam.coef_))

splines = utils.b_spline_basis(X, edge_knots=knot_edges, sparse=False)

splines_df = pd.DataFrame(splines, columns=[f'basis_{i}' for i in range(len(gam.coef_))])

splines_df.head(10)

	basis_0	basis_1	basis_2	basis_3	basis_4	basis_5
0	0.166667	0.666667	0.166667	0.000000	0.000000	0.000000
1	0.132997	0.661606	0.205334	0.000063	0.000000	0.000000
2	0.059688	0.594827	0.341426	0.004059	0.000000	0.000000
3	0.030095	0.518726	0.437481	0.013698	0.000000	0.000000
4	0.012374	0.428013	0.527144	0.032470	0.000000	0.000000
5	0.000000	0.040337	0.551431	0.399315	0.008917	0.000000
6	0.000000	0.018232	0.465467	0.492630	0.023671	0.000000
7	0.000000	0.011137	0.418489	0.535407	0.034967	0.000000
8	0.000000	0.000014	0.189310	0.664817	0.145859	0.000000
9	0.000000	0.000000	0.120914	0.656897	0.222015	0.000174

These spline features range the extent of the of covariate space \(X\) defining “partitions” of the space. The model “learns” to capture the shape of the outcome variable \(y\) by figuring out how to weight the different portion of this spline basis matrix i.e. the linear combination of this basis matrix with the derived coefficients is a model of our outcome variable.

Note how each row is 0 everywhere except within the columns that represent a partition of \(X\). Additionally each row sum to unity. These properties are important because they ensure that any weighted combination of this basis represents the outcome variable in a controlled and quite granular manner. The more splines we use the more control we have of the representation.

Plotting the Weighted Spline

ax = splines_df.dot(gam.coef_).plot(title='Weighted Splines', label='Weighted Combination of Spline Basis', figsize=(10, 6))
ax.set_ylabel("Acceleration")
ax.set_xlabel("Time Steps")
ax.legend();

In this manner we can see how the specification of a spline basis can help us model eccentric curves in an outcome space.

Next we’ll see how to more directly work with the specification of basis splines, passing these feature matrices into Bambi models.

Bayesian Splines with bambi

Under the hood Bambi makes use of the patsy package formula syntax to specify spline basis terms.

knots_6 = np.linspace(0, np.max(X), 6+2)[1:-1]
knots_10 = np.linspace(0, np.max(X), 10+2)[1:-1]
knots_15 = np.linspace(0, np.max(X), 15+2)[1:-1]

df = pd.DataFrame({'X': X.flatten(), 'y': y})
formula1 = 'bs(X, degree=0, knots=knots_6)'
formula2 = 'bs(X, degree=1, knots=knots_6, intercept=True)'
formula3 = 'bs(X, degree=3, knots=knots_6, intercept=True)'
formula4 = 'bs(X, degree=3, knots=knots_10, intercept=True)'
formula5 = 'bs(X, degree=3, knots=knots_15, intercept=True)'
model_spline1 = bmb.Model(f"y ~ {formula1}", df)
model_spline2 = bmb.Model(f"y ~ {formula2}", df)
model_spline3 = bmb.Model(f"y ~ {formula3}", df)
model_spline4 = bmb.Model(f"y ~ {formula4}", df)
model_spline5 = bmb.Model(f"y ~ {formula5}", df)
model_spline5

       Formula: y ~ bs(X, degree=3, knots=knots_15, intercept=True)
        Family: gaussian
          Link: mu = identity
  Observations: 133
        Priors: 
    target = mu
        Common-level effects
            Intercept ~ Normal(mu: -25.5459, sigma: 226.5005)
            bs(X, degree=3, knots=knots_15, intercept=True) ~ Normal(mu: [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
                0. 0. 0. 0. 0. 0. 0. 0.], sigma: [1209.5913 1309.9059 1516.6254  914.7421  803.2567
                526.1027  542.4563
              739.4915  590.0538  692.802   786.9096  707.6947  852.6529  792.6311
             1046.3175 1246.4855 1367.9865 1595.1739 1391.2518])
        
        Auxiliary parameters
            sigma ~ HalfStudentT(nu: 4.0, sigma: 48.14)

As you can see here we are specifying a range of different degrees of spline basis. The different degrees corresspond to the smoothness of the overlapping splines. The degree=0 splines mean we specify a piecewise constant basis i.e. 0 or 1 within each region of the partition. But we can add more degrees to see more flexible representations of the space.

model_spline5.build()
model_spline5.graph()

This will become clearer as we plot the various spline basis matrices below.

Plot the Spline Basis

The below functions extract the basis specification from each model and plots the basis design for an increasingly complex series of spline basis matrices.

def plot_spline_basis(basis, X, ax, title="Spline Basis"):
    df = (
        pd.DataFrame(basis)
        .assign(X=X)
        .melt("X", var_name="basis_idx", value_name="y")
    )


    for idx in df.basis_idx.unique():
        d = df[df.basis_idx == idx]
        ax.plot(d["X"], d["y"])
    
    ax.set_title(title)
    return ax

def plot_knots(knots, ax):
    for knot in knots:
        ax.axvline(knot, color="0.1", alpha=0.4)
    return ax



fig, axs = plt.subplots(5, 1, figsize=(9, 20))
axs = axs.flatten()
axs.flatten()
B1 = model_spline1.response_component.design.common[formula1]
plot_spline_basis(B1, df["X"].values, ax=axs[0], title="Piecewise Constant Basis")
plot_knots(knots_6, axs[0]);

B2 = model_spline2.response_component.design.common[formula2]
ax = plot_spline_basis(B2, df["X"].values, axs[1], 
title="Piecewise Linear Basis")
plot_knots(knots_6, axs[1]);

B3 = model_spline3.response_component.design.common[formula3]
ax = plot_spline_basis(B3, df["X"].values, axs[2], 
title="Cubic Spline Basis (6 Knots)")
plot_knots(knots_6, axs[2]);

B4 = model_spline4.response_component.design.common[formula4]
ax = plot_spline_basis(B4, df["X"].values, axs[3], 
title="Cubic Spline Basis (10 Knots)")
plot_knots(knots_10, axs[3]);


B5 = model_spline5.response_component.design.common[formula5]
ax = plot_spline_basis(B5, df["X"].values, axs[4], 
title="Cubic Spline Basis (15 Knots)")
plot_knots(knots_15, axs[4]);

Now that we’ve seen the nature of the modelling splines we’ll fit each model to the data and plot how the weighted splines matrices are able to represent the raw data.

Fit the Individual Spline Models

idata_spline1 = model_spline1.fit(random_seed=random_seed, idata_kwargs={"log_likelihood": True})

idata_spline2 = model_spline2.fit(random_seed=random_seed, idata_kwargs={"log_likelihood": True})

idata_spline3 = model_spline3.fit(random_seed=random_seed, idata_kwargs={"log_likelihood": True})

idata_spline4 = model_spline4.fit(random_seed=random_seed, idata_kwargs={"log_likelihood": True})

idata_spline5 = model_spline5.fit(random_seed=random_seed, idata_kwargs={"log_likelihood": True})

Plotting the fits

Plot the Weighted Mean

def plot_weighted_splines(B, idata, formula, ax, knots):
    posterior_stacked = az.extract(idata)
    wp = posterior_stacked[formula].mean("sample").values

    plot_spline_basis(B * wp.T, df["X"].values, ax)
    ax.plot(df.X.values, np.dot(B, wp.T), color="black", lw=3, label='Weighted Splines')
    plot_knots(knots, ax);
    ax.legend()



fig, axs = plt.subplots(5, 1, figsize=(10, 20))
axs = axs.flatten()
axs.flatten()

plot_weighted_splines(B1, idata_spline1, formula1, axs[0], knots_6)
plot_weighted_splines(B2, idata_spline2, formula2, axs[1], knots_6)
plot_weighted_splines(B3, idata_spline3, formula3, axs[2], knots_6)
plot_weighted_splines(B4, idata_spline4, formula4, axs[3], knots_10)
plot_weighted_splines(B5, idata_spline5, formula5, axs[4], knots_15)

Here we can see how the models with increasingly complex splines are more exactly able to fit the herky jerky trajectory of the outcome variable in each interval. The fewer the intervals, the less flexibility available to the model.

Compare Model Fits

As before we can evaluate these model fits and compare them based on leave-one-out cross validation scores and information theoretic complexity measures.

fig, axs = plt.subplots(1, 2, figsize=(10, 7))
axs = axs.flatten()
models_dict = {"piecewise_constant": idata_spline1, "piecewise_linear": idata_spline2, "cubic_bspline": idata_spline3, "cubic_bspline_10": idata_spline4, 
"cubic_bspline_15": idata_spline5}
df_compare = az.compare(models_dict)
az.plot_compare(df_compare, ax=axs[0])
az.plot_compare(az.compare(models_dict, 'waic'), ax=axs[1])
axs[1].set_yticklabels([])
df_compare

	rank	elpd_loo	p_loo	elpd_diff	weight	se	dse	warning	scale
cubic_bspline_10	0	-612.274608	11.462200	0.000000	9.030402e-01	9.689854	0.000000	True	log
cubic_bspline_15	1	-616.252712	15.566839	3.978104	2.408230e-14	9.833120	1.374420	True	log
cubic_bspline	2	-634.729956	8.698337	22.455348	5.894750e-03	8.897139	6.637243	True	log
piecewise_constant	3	-643.781042	6.981098	31.506433	5.164397e-02	9.770740	8.719803	False	log
piecewise_linear	4	-647.267410	6.170543	34.992802	3.942108e-02	7.902853	8.094122	False	log

Here we see that the extra complexity of using 15 splines leads to slightly worse performance measures than the less complex but seemingly adequate 10 splines.

new_data = pd.DataFrame({"X": np.linspace(df.X.min(), df.X.max(), num=500)})
    
model_spline4.predict(idata_spline4, data=new_data, 
kind='pps', inplace=True)

idata_spline4

Next we plot the posterior predictive distribution of our observed variable and compare against the observed data. Additionally we plot the 89th and 50% HDI.

ax = az.plot_hdi(new_data['X'], idata_spline4['posterior_predictive']['y'], fill_kwargs={'alpha': 0.2, 'color':'firebrick'}, hdi_prob=0.89, figsize=(10, 8))

az.plot_hdi(new_data['X'], idata_spline4['posterior_predictive']['y'], fill_kwargs={'alpha': 0.8, 'color':'firebrick'}, hdi_prob=0.5)

y_mean = idata_spline4['posterior_predictive']['y'].mean(dim=('chain', 'draw'))

ax.plot(new_data['X'], y_mean, label='Expected posterior predictive', color='k')
ax.set_xlabel("Time Point")
ax.set_ylabel("Acceleration")

ax.scatter(df['X'], df['y'], label='Observed Datapoints')

ax.legend()

ax.set_title("Posterior Predictive Distribution \n Based on 10 Knots");

This represents a good a clean model fit to the observed data using univariate spline smoothers. Next we’ll see another alternative approach to model this outcome variable using approximate gaussian processes.

Gaussian processes

The topic of gaussian processes is rich and detailed. Too rich to be fairly covered in this blog post, so we’ll just say that we’re using a method designed for function approximation that makes use of drawing samples from a multivariate normal distribution under a range of different covariance relationships.

These relationships can be somewhat intuitively interrogated by defining different combinations of covariance relationships with priors over the parameters governing the covariance of a sequence of points. For example consider the following parameterisations.

lengthscale = 3
sigma = 13
cov = sigma**2 * pm.gp.cov.ExpQuad(1, lengthscale)

X = np.linspace(0, 60, 200)[:, None]
K = cov(X).eval()

fig, ax = plt.subplots(figsize=(9, 7))
ax.plot(
    X,
    pm.draw(
        pm.MvNormal.dist(mu=np.zeros(len(K)), cov=K, shape=K.shape[0]), draws=10, random_seed=random_seed
    ).T,
)
plt.title(f"Samples from the GP prior \n lengthscale: {3}, sigma: {13}")
plt.ylabel("y")
plt.xlabel("X");

We’ve specified the range of X to reflect the support of the acceleration example and allowed the draws to be informed by a covariance function we have parameterised using the Exponentiated Quadratic kernel:

\[k(x, x') = \mathrm{exp}\left[ -\frac{(x - x')^2}{2 \ell^2} \right]\]

The patterns exhibited show a good range of “wiggliness” that they should be flexible enough to capture the shape of the acceleration, if we can calibrate the posterior of parameters against the observed data.

Priors on Gaussian Priors

Consider the following specification for the priors

fig, axs = plt.subplots(1, 2, figsize=(9, 6))
axs = axs.flatten()
axs[0].hist(pm.draw(pm.InverseGamma.dist(mu=1, sigma=1), 1000), ec='black', bins=30);
axs[0].set_title("Priors for Lengthscale \n in ExpQuad Kernel")
axs[1].hist(pm.draw(pm.Exponential.dist(lam=1), 1000), ec='black', bins=30);
axs[1].set_title("Priors for Amplitude \n in ExpQuad Kernel")

Text(0.5, 1.0, 'Priors for Amplitude \n in ExpQuad Kernel')

We use these to specify priors on the Hilbert space approximation of gaussian priors available in the Bambi package.

prior_hsgp = {
    "sigma": bmb.Prior("Exponential", lam=1), # amplitude
    "ell": bmb.Prior("InverseGamma", mu=1, sigma=1) # lengthscale
}

# This is the dictionary we pass to Bambi
priors = {
    "hsgp(X, m=10, c=1)": prior_hsgp,
    "sigma": bmb.Prior("HalfNormal", sigma=4)
}
model_hsgp = bmb.Model("y ~ 0 + hsgp(X, m=10, c=1)", df, priors=priors)
model_hsgp

       Formula: y ~ 0 + hsgp(X, m=10, c=1)
        Family: gaussian
          Link: mu = identity
  Observations: 133
        Priors: 
    target = mu
        HSGP contributions
            hsgp(X, m=10, c=1)
                cov: ExpQuad
                sigma ~ Exponential(lam: 1.0)
                ell ~ InverseGamma(mu: 1.0, sigma: 1.0)
        
        Auxiliary parameters
            sigma ~ HalfNormal(sigma: 4.0)

Here we’ve set the m=10 to determine the number of basis vectors used in the Hilbert space approximation. The idea differs in detail from the spline based approximations we’ve seen, but it’s perhaps useful to think of the process in the same vein.

idata = model_hsgp.fit(inference_method="nuts_numpyro",target_accept=0.95, random_seed=121195, 
idata_kwargs={"log_likelihood": True})
print(idata.sample_stats["diverging"].sum().to_numpy())

Compiling...

Compilation time = 0:00:01.012539

Sampling...

Sampling time = 0:00:01.792848

Transforming variables...

Transformation time = 0:00:00.183230

Computing Log Likelihood...

Log Likelihood time = 0:00:00.151209

This model fits and the sampling seems to have worked well.

az.plot_trace(idata, backend_kwargs={"layout": "constrained"}, figsize=(9, 15));

The lengthscale and sigma parameters we have learned by calibrating our priors against the data. The degree to which these parameters are meaningful depend a little on how familar you are with covariance matrix kernels and their properties, so we won’t dwell on the point here.

az.summary(idata, var_names=['hsgp(X, m=10, c=1)_ell', 'hsgp(X, m=10, c=1)_sigma', 'y_sigma', 'hsgp(X, m=10, c=1)_weights'])

	mean	sd	hdi_3%	hdi_97%	mcse_mean	mcse_sd	ess_bulk	ess_tail	r_hat
hsgp(X, m=10, c=1)_ell	3.301	0.666	2.099	4.567	0.016	0.011	1763.0	1815.0	1.0
hsgp(X, m=10, c=1)_sigma	23.675	2.679	18.925	28.847	0.069	0.049	1497.0	2167.0	1.0
y_sigma	20.589	1.121	18.590	22.751	0.019	0.014	3440.0	2485.0	1.0
hsgp(X, m=10, c=1)_weights[0]	-131.588	13.784	-157.019	-105.189	0.216	0.154	4071.0	3020.0	1.0
hsgp(X, m=10, c=1)_weights[1]	-94.391	18.010	-125.921	-58.020	0.363	0.256	2500.0	2876.0	1.0
hsgp(X, m=10, c=1)_weights[2]	106.121	20.717	69.627	147.247	0.458	0.324	2046.0	2563.0	1.0
hsgp(X, m=10, c=1)_weights[3]	149.289	22.154	106.508	188.107	0.521	0.368	1807.0	2467.0	1.0
hsgp(X, m=10, c=1)_weights[4]	-83.399	22.728	-124.812	-40.406	0.539	0.383	1776.0	2517.0	1.0
hsgp(X, m=10, c=1)_weights[5]	-125.322	23.273	-169.346	-82.187	0.531	0.377	1917.0	2940.0	1.0
hsgp(X, m=10, c=1)_weights[6]	37.835	21.153	-1.712	77.574	0.512	0.362	1713.0	2577.0	1.0
hsgp(X, m=10, c=1)_weights[7]	94.574	20.745	53.167	132.182	0.423	0.299	2393.0	3105.0	1.0
hsgp(X, m=10, c=1)_weights[8]	-1.494	17.184	-31.835	33.063	0.361	0.255	2264.0	2853.0	1.0
hsgp(X, m=10, c=1)_weights[9]	-45.442	15.955	-76.649	-16.636	0.289	0.208	3041.0	3183.0	1.0

But again we can sample from the posterior predictive distribution of the outcome variable

model_hsgp.predict(idata, data=new_data, 
kind='pps', inplace=True)

idata

and plot the model fit to see if it can recover the observed data.

ax = az.plot_hdi(new_data['X'], idata['posterior_predictive']['y'], fill_kwargs={'alpha': 0.2, 'color':'firebrick'}, figsize=(9, 8))

az.plot_hdi(new_data['X'], idata['posterior_predictive']['y'], fill_kwargs={'alpha': 0.8, 'color':'firebrick'}, hdi_prob=0.5)

y_mean = idata['posterior_predictive']['y'].mean(dim=('chain', 'draw'))

ax.plot(new_data['X'], y_mean, label='Expected posterior predictive', color='k')

ax.scatter(df['X'], df['y'], label='Observed Datapoints')

ax.legend()

ax.set_title("Posterior Predictive Distribution \n Based on HSGP approximation")

Text(0.5, 1.0, 'Posterior Predictive Distribution \n Based on HSGP approximation')

And we can compare versus the spline models to see that by the aggregate performance measures our HSGP model seems to come out on top.

models_dict = {"piecewise_constant": idata_spline1, "piecewise_linear": idata_spline2, "cubic_bspline": idata_spline3, "cubic_bspline_10": idata_spline4, 
"cubic_bspline_15": idata_spline5, 'hsgp': idata}
df_compare = az.compare(models_dict)
df_compare

	rank	elpd_loo	p_loo	elpd_diff	weight	se	dse	warning	scale
hsgp	0	-608.819770	9.958827	0.000000	0.873213	12.009698	0.000000	False	log
cubic_bspline_10	1	-612.278059	11.465651	3.458289	0.000000	9.688433	2.962396	True	log
cubic_bspline_15	2	-616.252712	15.566839	7.432943	0.000000	9.833120	3.457579	True	log
cubic_bspline	3	-634.729956	8.698337	25.910186	0.000000	8.897139	7.686648	True	log
piecewise_constant	4	-643.781042	6.981098	34.961272	0.069506	9.770740	9.490817	False	log
piecewise_linear	5	-647.267410	6.170543	38.447641	0.057282	7.902853	9.200728	False	log

Recap

So far we’ve seen how we can use splines and gauassian processes to model highly eccentric functional relationships where the function could be approximated with univariate smoothing routine. Next we’ll show how to use hierarchical modelling over spline fits to extract insight into the data generating process over a family of curves. In particular we’ll focus on the development of insurance loss curves.

Insurance Loss Curves: Hierarchical Spline Models

We draw on car insurance losses data set discussed in Mick Cooney’s Stan case-study, but we simplify things for ourselves considerably by focusing on two types of loss and ensuring that each year under consideration has equal observations of the accruing losses.

loss_df = pd.read_csv('ppauto_pos.csv')
loss_df = loss_df[(loss_df['GRCODE'].isin([43, 353])) & (loss_df['DevelopmentYear'] < 1998)]

loss_df = loss_df[['GRCODE', 'AccidentYear', 'DevelopmentYear', 'DevelopmentLag', 'EarnedPremDIR_B', 'CumPaidLoss_B']]


loss_df.columns = ['grcode', 'acc_year', 'dev_year', 'dev_lag', 'premium', 'cum_loss']
loss_df['lr'] = loss_df['cum_loss'] / loss_df['premium']
loss_df = loss_df[(loss_df['acc_year'] <= 1992) & (loss_df['dev_lag'] <= 6)].reset_index(drop=True)

loss_df['year_code'] = loss_df['acc_year'].astype(str) + '_' + loss_df['grcode'].astype(str)
loss_df.sort_values(by=['year_code', 'acc_year', 'dev_lag'], inplace=True)
loss_df['standardised_premium'] = (loss_df.assign(mean_premium = np.mean(loss_df['premium']))
.assign(std_premium = np.std(loss_df['premium']))
.apply(lambda x: (x['mean_premium'] - x['premium']) /x['std_premium'], axis=1)
)

loss_df.head(12)

	grcode	acc_year	dev_year	dev_lag	premium	cum_loss	lr	year_code	standardised_premium
30	353	1988	1988	1	18793	4339	0.230884	1988_353	-0.347453
31	353	1988	1989	2	18793	9617	0.511733	1988_353	-0.347453
32	353	1988	1990	3	18793	11584	0.616400	1988_353	-0.347453
33	353	1988	1991	4	18793	12001	0.638589	1988_353	-0.347453
34	353	1988	1992	5	18793	12640	0.672591	1988_353	-0.347453
35	353	1988	1993	6	18793	12966	0.689938	1988_353	-0.347453
0	43	1988	1988	1	957	133	0.138976	1988_43	1.722344
1	43	1988	1989	2	957	333	0.347962	1988_43	1.722344
2	43	1988	1990	3	957	431	0.450366	1988_43	1.722344
3	43	1988	1991	4	957	570	0.595611	1988_43	1.722344
4	43	1988	1992	5	957	615	0.642633	1988_43	1.722344
5	43	1988	1993	6	957	615	0.642633	1988_43	1.722344

Plot the Loss Curves

Here we have plotted the developing loss curves from two different coded insurance products.

pivot = loss_df.pivot(index=['dev_lag'], columns=['grcode', 'acc_year'], values='lr')
fig, axs = plt.subplots(1, 2, figsize=(9, 7))
pivot.plot(figsize=(10, 6), ax=axs[0])
axs[0].set_title("Loss Ratios by Year");
for c in pivot.columns:
    if 43 in c:
        color='red'
    else: 
        color='grey'
    axs[1].plot(pivot[c], color=color, label=c)
axs[1].legend()
axs[1].set_title("Loss Ratio by Group")

Text(0.5, 1.0, 'Loss Ratio by Group')

We want to model these curves collectively as instances of draws from a distribution of loss curves. To do so we will specify a PyMC hierarchical (mixed) spline model. To do so we will have a spline basis for the global hyper parameters beta_g and the individual parameters for each curve. Here we define a convenience function to generate the basis splines.

from patsy import bs as bs_patsy, dmatrix

def make_basis_splines(num_knots=3, max_dev=7):
    knot_list = np.linspace(0, max_dev, num_knots+2)[1:-1]
    dev_periods = np.arange(1, max_dev, 1)

    Bi = dmatrix(
        "bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True) - 1",
        {"dev_periods": dev_periods, "knots": knot_list},
    )

    Bg = dmatrix(
        "bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True) - 1",
        {"dev_periods": dev_periods, "knots": knot_list})


    return Bi, Bg

Bi, Bg = make_basis_splines()
Bg

DesignMatrix with shape (6, 7)
  Columns:
    ['bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True)[0]',
     'bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True)[1]',
     'bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True)[2]',
     'bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True)[3]',
     'bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True)[4]',
     'bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True)[5]',
     'bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True)[6]']
  Terms:
    'bs_patsy(dev_periods, knots=knots, degree=3, include_intercept=True)' (columns 0:7)
  (to view full data, use np.asarray(this_obj))

Next we specify a model maker function to create the various pooled, unpooled and hierarhical (mixed) models of the insurance curve data. Note that even though we’re specifying a hierarhical model we have not specified a hierarchy over the insurance codes, instead we have added this as a “fixed” effect feature into our regression model. The idea is that this fixed effect will capture the differences in expected baseline per group code of insurance product.

def make_model(loss_df, num_knots=3, max_dev=7, model_type='mixed'):
    Bi, Bg = make_basis_splines(num_knots, max_dev)
    observed = loss_df['lr'].values
    uniques, unique_codes = pd.factorize(loss_df['year_code'])
    coords= {'years': unique_codes, 
            'splines': list(range(Bi.shape[1])) ,
            'measurement': list(range(6)), 
            'obs': uniques
            }

    with pm.Model(coords=coords) as sp_insur:
        basis_g = pm.MutableData('Bg', np.asfortranarray(Bg))

        tau = pm.HalfCauchy('tau', 1)
        ## Global Hierarchical Spline Terms
        beta_g = pm.Normal("beta_g", mu=0, sigma=tau, 
        dims='splines')
        mu_g = pm.Deterministic("mu_g", pm.math.dot(basis_g, beta_g), dims='measurement')

        ## Individual or Year Specific Spline Modifications
        if model_type in ['mixed', 'unpooled']:
            basis_i = pm.MutableData('Bi', np.asfortranarray(Bi))
            beta = pm.Normal("beta", mu=0, sigma=tau, dims=('splines', 'years'))
            mui = pm.Deterministic("mui", pm.math.dot(basis_i, beta), dims=('measurement', 'years'))
        
        ## Features
        prem = pm.MutableData('prem', loss_df['standardised_premium'].values)
        grcode = pm.MutableData('grcode', loss_df['grcode'] == 43)

        beta_prem = pm.Normal('beta_prem', 0, 1)
        beta_grcode = pm.Normal('beta_grcode', 0, 1)
        mu_prem = beta_prem*prem
        mu_grcode = beta_grcode*grcode

        ## Likelihood
        sigma = pm.TruncatedNormal("sigma", 1, lower=0.05)
        if model_type == 'mixed':
            mu = pm.Deterministic('mu',  mu_grcode + mu_prem + (mu_g.T + mui.T).ravel(), dims='obs')
            lr_likelihood = pm.Normal("lr", mu, sigma, observed=observed, dims=('obs'))
        elif model_type == 'pooled': 
             lr_likelihood = pm.Normal("lr",  np.repeat(mu_g, len(unique_codes)), sigma, observed=observed, dims='obs')
        elif model_type == 'unpooled':
            lr_likelihood = pm.Normal("lr",  mui.T.ravel(), sigma, observed=observed, dims=('obs'))


        ## Sampling
        idata_sp_insur = pm.sample(2000, return_inferencedata=True, target_accept=.99,
        idata_kwargs={"log_likelihood": True})
        idata_sp_insur = pm.sample_posterior_predictive(
            idata_sp_insur,extend_inferencedata=True)

    return idata_sp_insur, sp_insur


idata_sp_insur_unpooled, sp_insur_unpooled = make_model(loss_df, model_type='unpooled')
idata_sp_insur_pooled, sp_insur_pooled = make_model(loss_df, model_type='pooled')
idata_sp_insur_mixed, sp_insur_mixed = make_model(loss_df, model_type='mixed')

The model structure can be seen more clearly in this graph

pm.model_to_graphviz(sp_insur_mixed)

We can extract the effect of the differences grcodes and examine the baseline and annual spline related coefficients.

summary = az.summary(idata_sp_insur_mixed, var_names=['beta_grcode', 'beta_prem', 'beta_g', 'beta'])

summary

	mean	sd	hdi_3%	hdi_97%	mcse_mean	mcse_sd	ess_bulk	ess_tail	r_hat
beta_grcode	0.333	0.075	0.191	0.474	0.001	0.001	4085.0	5187.0	1.0
beta_prem	-0.056	0.038	-0.123	0.018	0.001	0.000	4188.0	4896.0	1.0
beta_g[0]	0.077	0.090	-0.082	0.257	0.002	0.001	2977.0	3549.0	1.0
beta_g[1]	0.056	0.175	-0.285	0.373	0.002	0.002	6980.0	6019.0	1.0
beta_g[2]	0.434	0.135	0.193	0.700	0.002	0.001	5982.0	6033.0	1.0
...	...	...	...	...	...	...	...	...	...
beta[6, 1990_43]	0.551	0.104	0.356	0.748	0.002	0.001	3459.0	5044.0	1.0
beta[6, 1991_353]	-0.013	0.099	-0.199	0.178	0.002	0.001	3405.0	4314.0	1.0
beta[6, 1991_43]	0.107	0.105	-0.099	0.297	0.002	0.001	3627.0	4773.0	1.0
beta[6, 1992_353]	-0.028	0.100	-0.215	0.158	0.002	0.001	3498.0	4643.0	1.0
beta[6, 1992_43]	-0.069	0.125	-0.298	0.166	0.002	0.001	3888.0	5137.0	1.0

79 rows × 9 columns

Again we can compare the performance metrics of the various models.

compare_df = az.compare({'unpooled': idata_sp_insur_unpooled, 
            'pooled': idata_sp_insur_pooled, 
            'mixed': idata_sp_insur_mixed})

az.plot_compare(compare_df)
compare_df

	rank	elpd_loo	p_loo	elpd_diff	weight	se	dse	warning	scale
mixed	0	62.790045	35.822036	0.000000	1.000000e+00	2.360433	0.000000	True	log
unpooled	1	43.536513	44.579069	19.253532	1.479822e-11	1.409566	1.698917	True	log
pooled	2	-9.219961	5.868017	72.010006	0.000000e+00	5.376135	4.684697	False	log

Plot the Posterior Predictive Checks

We can check how well the model can recapture the observed data.

def plot_ppc_splines(idata):
    fig, axs = plt.subplots(5, 2, figsize=(9, 20), sharey=True)
    axs = axs.flatten()
    dev_periods = np.arange(1, 7, 1)
    uniques, unique_codes = pd.factorize(loss_df['year_code'])
    for y, c in zip(unique_codes, range(10)):
        az.plot_hdi(dev_periods, idata['posterior_predictive']['lr'].sel(obs=c), color='firebrick', ax=axs[c], fill_kwargs={'alpha': 0.2}, hdi_prob=.89)
        az.plot_hdi(dev_periods, idata['posterior_predictive']['lr'].sel(obs=c), color='firebrick', ax=axs[c], hdi_prob=0.5)
        axs[c].scatter(dev_periods, loss_df[(loss_df['year_code'] == y)]['lr'], color='k', label='Actual Loss Ratio')
        axs[c].plot(dev_periods, loss_df[(loss_df['year_code'] == y)]['lr'], color='k')
        axs[c].set_title(f"PPC 89% and 50% HDI: {y}")
        axs[c].set_ylabel("Loss Ratio")
        axs[c].legend();

plot_ppc_splines(idata_sp_insur_mixed)

Plot the Hierarchical Components

In the following plot we show similarly how to recapture the observed data, but additionally we decompose the structure of the model in each case and extract baseline forecasts which would be our guide to future loss-ratio development curves in lieu of any other information.

mu_g = idata_sp_insur_mixed.posterior.stack(draws=("chain", "draw"))["mu_g"]

mu_i = idata_sp_insur_mixed.posterior.stack(draws=("chain", "draw"))["mui"]

mu  = idata_sp_insur_mixed.posterior.stack(draws=("chain", "draw"))['mu']

beta_grcode = idata_sp_insur_mixed.posterior.stack(draws=("chain", "draw"))['beta_grcode']

dev_periods = np.arange(1, 7, 1)
uniques, unique_codes = pd.factorize(loss_df['year_code'])

mosaic = """
         AB
         CD
         EF
         GH
         IJ
         KK
"""
fig, axs = plt.subplot_mosaic(mosaic, sharey=True, 
figsize=(9, 20))
axs = [axs[k] for k in axs.keys()] 

mu_g_mean = mu_g.mean(dim='draws')
for y, c in zip(unique_codes, range(10)):
    group_effect = 0
    if '43' in y: 
        group_effect = beta_grcode.mean().item()
    mu_i_mean = mu_i.sel(years=y).mean(dim='draws')
    axs[c].plot(dev_periods, group_effect + mu_g_mean.values + mu_i_mean.values, label='Combined + E(grp_effect)', color='purple', linewidth=3.5)
    axs[c].plot(dev_periods, group_effect + mu_g_mean.values, label='E(Hierarchical Baseline)', color='red', linestyle='--')
    axs[c].plot(dev_periods,  group_effect + mu_i_mean.values, label='E(Year Specific Adjustment term)', color='blue', linestyle='--')
    axs[c].scatter(dev_periods, loss_df[(loss_df['year_code'] == y)]['lr'], color='k', label='Actual Loss Ratio')
    az.plot_hdi(dev_periods,mu.sel(obs=c).T  , ax=axs[c], color='firebrick', fill_kwargs={'alpha': 0.2})
    az.plot_hdi(dev_periods, mu.sel(obs=c).T , ax=axs[c], color='firebrick', fill_kwargs={'alpha': 0.5}, hdi_prob=.50)
    axs[c].set_title(f"Components for Year {y}")
    axs[c].set_ylabel("Loss Ratio")
    if (c == 0) or (c == 1):
         axs[c].legend()

axs[10].plot(dev_periods, mu_g_mean.values, label='E(Hierarchical Baseline)', color='black')
axs[10].plot(dev_periods, mu_g_mean.values + group_effect, label='E(Hierarchical Baseline) + E(grp_effect)', color='black', linestyle='--')
az.plot_hdi(dev_periods, mu_g.T.values, color='slateblue', ax=axs[10], fill_kwargs={'alpha': 0.2})
az.plot_hdi(dev_periods, mu_g.T.values + group_effect, color='magenta', ax=axs[10], fill_kwargs={'alpha': 0.2})
az.plot_hdi(dev_periods, mu_g.T.values, color='slateblue', ax=axs[10], hdi_prob=.5)
az.plot_hdi(dev_periods, mu_g.T.values  + group_effect, color='magenta', ax=axs[10], hdi_prob=.5)
axs[10].set_title("Baseline Forecast Loss Ratio")
axs[10].legend();

This plot is a bit clunky, because we’re mixing expectations and posterior distributions over the parameters. The point is just to highlight the “compositional” structure of our model. A better way to interrogate the implications of the model is to “push” forward different data through the posterior predictive distribution and derive a kind of ceteris paribus rule for accrual of losses.

with sp_insur_mixed: 
    pm.set_data({'grcode': np.ones(len(loss_df)), 
    })
    idata_43 = pm.sample_posterior_predictive(idata_sp_insur_mixed, var_names=['lr'], extend_inferencedata =True)

with sp_insur_mixed: 
    pm.set_data({'grcode': np.zeros(len(loss_df))})
    idata_353 = pm.sample_posterior_predictive(idata_sp_insur_mixed, var_names=['lr'], extend_inferencedata=True)

idata_353

Sampling: [lr]

100.00% [8000/8000 00:00<00:00]

Sampling: [lr]

100.00% [8000/8000 00:00<00:00]

Even here though we want to average the curves over the specific years in the data and abstract a view of the model implications under different counterfactual settings. Or put another way we a view of the average development curve between years not within a year. Here we define a helper function to effect this step.

def get_posterior_predictive_curve(idata, prem=2, grcode=1):
    weighted_splines = [np.dot(np.asfortranarray(Bi), az.extract(idata['posterior']['beta'])['beta'].values[:, :, i]) for i in range(1000)]

    weighted_splines_1 = [np.dot(np.asfortranarray(Bg), az.extract(idata['posterior']['beta_g'])['beta_g'].values[:, i]) for i in range(1000)]

    beta_grcode = az.extract(idata['posterior']['beta_grcode'])['beta_grcode']

    beta_prem = az.extract(idata['posterior']['beta_prem'])['beta_prem']
    df1 = pd.DataFrame([beta_prem.values[i]*prem + beta_grcode.values[i]*grcode for i in range(1000)]).T

    ## Crucial step averaging over the years to get
    ## a view of the development period
    df = pd.concat([pd.DataFrame(weighted_splines_1[i].T + weighted_splines[i].T).mean() for i in range(1000)], axis=1)

    df = df1.iloc[0] + df

    return df

pred_df_1 = get_posterior_predictive_curve(idata_43, prem=1, grcode=1)

pred_df_0 = get_posterior_predictive_curve(idata_353, prem=1, grcode=0) 

fig, ax = plt.subplots(figsize=(9, 7), sharey=True)


ax.plot(pred_df_1, color='firebrick', alpha=0.1)
ax.plot(pred_df_0, color='slateblue', alpha=0.2);
ax.plot(pred_df_0.mean(axis=1), linestyle='-', color='k', linewidth=4, label='grcode 353 prem 1')

ax.plot(pred_df_1.mean(axis=1), linestyle='--', color='grey', linewidth=4, label='grcode 43 prem 1')

ax.set_title("Posterior Samples of the Trajectories \n Under different Counterfactual settings")
ax.set_ylabel("Loss Ratio")
ax.set_xlabel("Development Period")
ax.legend();

Multiple Smoother Regression Models

To demonstrate how spline modelling can be further adapting to the multiple regression like cases we use the PISA data set discussed in this case study from Michael Clark.

The data set has been constructed using average Science scores by country from the Programme for International Student Assessment (PISA) 2006, along with GNI per capita (Purchasing Power Parity, 2005 dollars), Educational Index, Health Index, and Human Development Index from UN data. We want to model the overall outcome score as a function of these broad demographic features.

pisa_df = pd.read_csv("https://raw.githubusercontent.com/m-clark/generalized-additive-models/master/data/pisasci2006.csv")

pisa_df

	Country	Overall	Issues	Explain	Evidence	Interest	Support	Income	Health	Edu	HDI
0	Albania	NaN	NaN	NaN	NaN	NaN	NaN	0.599	0.886	0.716	0.724
1	Argentina	391.0	395.0	386.0	385.0	567.0	506.0	0.678	0.868	0.786	0.773
2	Australia	527.0	535.0	520.0	531.0	465.0	487.0	0.826	0.965	0.978	0.920
3	Austria	511.0	505.0	516.0	505.0	507.0	515.0	0.835	0.944	0.824	0.866
4	Azerbaijan	382.0	353.0	412.0	344.0	612.0	542.0	0.566	0.780	NaN	NaN
...	...	...	...	...	...	...	...	...	...	...	...
60	Turkey	424.0	427.0	423.0	417.0	540.0	563.0	0.679	0.828	0.562	0.681
61	United Arab Emirates	NaN	NaN	NaN	NaN	NaN	NaN	0.909	0.878	0.686	0.818
62	United Kingdom	515.0	514.0	517.0	514.0	464.0	470.0	0.833	0.934	0.798	0.853
63	United States	489.0	492.0	486.0	489.0	480.0	490.0	0.872	0.911	0.930	0.904
64	Uruguay	428.0	429.0	423.0	429.0	567.0	510.0	0.658	0.884	0.740	0.755

65 rows × 11 columns

The relationships displayed between each of these measures is not obviously linear, and as such could plausibly benefit from being modelled with splines.

g = sns.pairplot(pisa_df[['Overall', 'Income', 'Support', 'Health', 'Edu']],  kind="reg")
g.fig.suptitle("Pair Plot of Complex Relations", y=1.05);

We define two basic models for contrasting. Note here how we have to define a seperate spline basis for each of the covariates.

formula = "Overall ~ Income + Edu + Health"
base_model = bmb.Model(formula, pisa_df, dropna=True)

knots_income = np.linspace(np.min(pisa_df['Income']), np.max(pisa_df['Income']), 5+2)[1:-1]

knots_edu = np.linspace(np.min(pisa_df['Edu']), np.max(pisa_df['Edu']), 5+2)[1:-1]

knots_health = np.linspace(np.min(pisa_df['Health']), np.max(pisa_df['Health']), 5+2)[1:-1]

formula_spline = """Overall ~ bs(Income, degree=3, knots=knots_income) + bs(Edu, degree=3, knots=knots_edu) + bs(Health, degree=3, knots=knots_health)"""

spline_model = bmb.Model(formula_spline, pisa_df, dropna=True)

spline_model

Automatically removing 13/65 rows from the dataset.

Automatically removing 13/65 rows from the dataset.

       Formula: Overall ~ bs(Income, degree=3, knots=knots_income) + bs(Edu, degree=3, knots=knots_edu) + bs(Health, degree=3, knots=knots_health)
        Family: gaussian
          Link: mu = identity
  Observations: 52
        Priors: 
    target = mu
        Common-level effects
            Intercept ~ Normal(mu: 471.1538, sigma: 464.5641)
            bs(Income, degree=3, knots=knots_income) ~ Normal(mu: [0. 0. 0. 0. 0. 0. 0. 0.], sigma:
                [2700.1849 1474.917   742.9386  517.2898  651.4592  515.7397  867.8945
              876.0978])
            bs(Edu, degree=3, knots=knots_edu) ~ Normal(mu: [0. 0. 0. 0. 0. 0. 0. 0.], sigma: [1120.7928
                920.8213  702.8955  537.103   560.453   582.8883 1023.4867
              680.9089])
            bs(Health, degree=3, knots=knots_health) ~ Normal(mu: [0. 0. 0. 0. 0. 0. 0. 0.], sigma:
                [1465.0684 3939.0308  690.5161  531.6981  655.7589  552.7209  642.9461
              925.9747])
        
        Auxiliary parameters
            sigma ~ HalfStudentT(nu: 4.0, sigma: 53.4654)

spline_model.build()
spline_model.graph()

base_idata = base_model.fit(random_seed=random_seed, idata_kwargs={"log_likelihood": True})
spline_idata = spline_model.fit(random_seed=random_seed, idata_kwargs={"log_likelihood": True}, target_accept=.95)

Auto-assigning NUTS sampler...

Initializing NUTS using jitter+adapt_diag...

Multiprocess sampling (4 chains in 4 jobs)

NUTS: [Overall_sigma, Intercept, Income, Edu, Health]

100.00% [8000/8000 00:02<00:00 Sampling 4 chains, 0 divergences]

Sampling 4 chains for 1_000 tune and 1_000 draw iterations (4_000 + 4_000 draws total) took 2 seconds.

Auto-assigning NUTS sampler...

Initializing NUTS using jitter+adapt_diag...

Multiprocess sampling (4 chains in 4 jobs)

NUTS: [Overall_sigma, Intercept, bs(Income, degree=3, knots=knots_income), bs(Edu, degree=3, knots=knots_edu), bs(Health, degree=3, knots=knots_health)]

100.00% [8000/8000 00:27<00:00 Sampling 4 chains, 0 divergences]

Sampling 4 chains for 1_000 tune and 1_000 draw iterations (4_000 + 4_000 draws total) took 28 seconds.

We can compare the simple regression approach to the spline based regression in the now usual way.

compare_df = az.compare({'spline': spline_idata, 'raw': base_idata})

compare_df

	rank	elpd_loo	p_loo	elpd_diff	weight	se	dse	warning	scale
spline	0	-254.101306	20.218465	0.000000	0.608049	4.724572	0.000000	True	log
raw	1	-263.061455	9.617888	8.960149	0.391951	10.083921	9.924089	True	log

The coefficients comparisons are harder

az.summary(base_idata)

	mean	sd	hdi_3%	hdi_97%	mcse_mean	mcse_sd	ess_bulk	ess_tail	r_hat
Intercept	118.438	81.996	-41.166	266.033	1.522	1.086	2915.0	2986.0	1.0
Income	181.955	89.547	17.413	353.598	1.913	1.353	2196.0	2432.0	1.0
Edu	234.125	59.153	120.739	340.343	1.017	0.740	3414.0	2518.0	1.0
Health	30.395	140.966	-223.699	298.258	2.852	2.153	2450.0	2598.0	1.0
Overall_sigma	34.234	3.620	27.882	41.175	0.060	0.043	3757.0	2879.0	1.0

since it’s less clear what the spline coefficient terms mean.

az.summary(spline_idata)

	mean	sd	hdi_3%	hdi_97%	mcse_mean	mcse_sd	ess_bulk	ess_tail	r_hat
Intercept	225.689	205.515	-133.222	630.021	6.036	4.292	1163.0	1653.0	1.0
bs(Income, degree=3, knots=knots_income)[0]	-355.159	1613.343	-3460.788	2642.535	56.357	39.865	819.0	1504.0	1.0
bs(Income, degree=3, knots=knots_income)[1]	194.905	122.719	-42.478	423.960	2.343	1.728	2748.0	2719.0	1.0
bs(Income, degree=3, knots=knots_income)[2]	48.748	123.638	-182.012	272.863	4.065	2.875	923.0	1431.0	1.0
bs(Income, degree=3, knots=knots_income)[3]	68.533	98.242	-118.858	250.496	3.369	2.383	853.0	1416.0	1.0
bs(Income, degree=3, knots=knots_income)[4]	210.290	103.892	17.347	406.753	3.626	2.565	825.0	1404.0	1.0
bs(Income, degree=3, knots=knots_income)[5]	222.546	105.553	31.163	427.878	3.589	2.539	866.0	1442.0	1.0
bs(Income, degree=3, knots=knots_income)[6]	202.395	105.402	0.453	394.110	3.629	2.567	846.0	1507.0	1.0
bs(Income, degree=3, knots=knots_income)[7]	81.592	103.313	-106.154	279.937	3.641	2.575	805.0	1390.0	1.0
bs(Edu, degree=3, knots=knots_edu)[0]	145.558	287.518	-410.399	654.146	8.525	6.029	1142.0	1509.0	1.0
bs(Edu, degree=3, knots=knots_edu)[1]	47.574	193.029	-323.504	392.406	5.595	3.957	1197.0	1656.0	1.0
bs(Edu, degree=3, knots=knots_edu)[2]	71.806	211.848	-312.889	469.050	6.246	4.417	1156.0	1647.0	1.0
bs(Edu, degree=3, knots=knots_edu)[3]	147.390	200.622	-224.182	518.214	5.972	4.224	1134.0	1603.0	1.0
bs(Edu, degree=3, knots=knots_edu)[4]	70.472	203.119	-314.815	427.662	6.018	4.256	1144.0	1535.0	1.0
bs(Edu, degree=3, knots=knots_edu)[5]	121.297	202.750	-264.164	485.547	5.998	4.243	1147.0	1575.0	1.0
bs(Edu, degree=3, knots=knots_edu)[6]	78.372	206.089	-296.582	457.669	6.012	4.252	1178.0	1713.0	1.0
bs(Edu, degree=3, knots=knots_edu)[7]	124.157	203.070	-258.062	490.727	6.008	4.249	1148.0	1532.0	1.0
bs(Health, degree=3, knots=knots_health)[0]	235.831	885.656	-1490.479	1862.011	31.104	22.002	813.0	1482.0	1.0
bs(Health, degree=3, knots=knots_health)[1]	162.805	389.636	-552.198	913.839	7.756	6.545	2529.0	2422.0	1.0
bs(Health, degree=3, knots=knots_health)[2]	14.967	120.574	-202.130	248.327	3.854	2.726	979.0	1657.0	1.0
bs(Health, degree=3, knots=knots_health)[3]	25.213	100.498	-170.867	204.686	3.411	2.413	871.0	1555.0	1.0
bs(Health, degree=3, knots=knots_health)[4]	-21.315	102.473	-223.005	158.665	3.577	2.530	822.0	1468.0	1.0
bs(Health, degree=3, knots=knots_health)[5]	-19.449	101.418	-213.704	165.809	3.541	2.505	823.0	1444.0	1.0
bs(Health, degree=3, knots=knots_health)[6]	-55.820	100.895	-252.216	126.193	3.547	2.509	811.0	1347.0	1.0
bs(Health, degree=3, knots=knots_health)[7]	0.840	101.892	-196.002	186.216	3.543	2.506	828.0	1469.0	1.0
Overall_sigma	24.766	3.303	18.816	30.884	0.085	0.060	1491.0	2068.0	1.0

We can also check that the model seems to recover the observed data well.

### Plotting Posterior Predictive Checks

base_model.predict(base_idata, kind='pps')
spline_model.predict(spline_idata, kind='pps')

fig, axs = plt.subplots(2, 1, figsize=(8, 12))
axs = axs.flatten()
az.plot_ppc(base_idata, ax=axs[0])
az.plot_ppc(spline_idata, ax=axs[1]);
axs[0].set_xlabel('')
axs[1].set_title("PPC: Spline Model");
axs[0].set_title("PPC: Regression Model");

Next we’ll dig into the spline basis features and decompose the predicted outcome and show how the outcome varies for levels in the inputs variables. The Bambi interpret package offers some functionality to assess the conditional predictions for varying values of the input variables.

Model Interpretability Plots in Bambi

fig, axs = plt.subplots(3, 1, figsize=(9, 10))
axs = axs.flatten()
bmb.interpret.plot_predictions(
    spline_model, spline_idata, "Income", ax=axs[0]
)

bmb.interpret.plot_predictions(
    spline_model, spline_idata, "Edu", ax=axs[1]
)

bmb.interpret.plot_predictions(
    spline_model, spline_idata, "Health", ax=axs[2]
);
axs[0].set_title("Predictions over the Range of Observed Covariate Values");

We can pull these types of values out into a table. Note here how we ask for predictions based on varying values of the Edu and Income variables where we keep the Health variable fixed at the mean value.

summary_df = bmb.interpret.predictions(
    spline_model,
    spline_idata,
    conditional={
        "Edu": list(np.linspace(0.5, 1, 10)),
        "Income": list(np.linspace(0.4, 1, 10)),
        },
)
summary_df

	Edu	Income	Health	estimate	lower_3.0%	upper_97.0%
0	0.5	0.400000	0.885672	-29.666186	-1954.336608	2094.868474
1	0.5	0.466667	0.885672	-228.798806	-2085.499823	1522.448806
2	0.5	0.533333	0.885672	-21.580853	-1613.700514	1468.840651
3	0.5	0.600000	0.885672	-57.471006	-1597.034080	1581.365824
4	0.5	0.666667	0.885672	-42.923777	-1539.633756	1595.098933
...	...	...	...	...	...	...
95	1.0	0.733333	0.885672	514.862194	438.494673	587.329605
96	1.0	0.800000	0.885672	568.402239	474.848846	663.243239
97	1.0	0.866667	0.885672	566.889704	479.214581	657.933780
98	1.0	0.933333	0.885672	289.886165	85.237681	495.492796
99	1.0	1.000000	0.885672	-1001.019928	-2494.557415	614.522981

100 rows × 6 columns

We can then plot these results based on the conditional realisations and see some interesting behaviour at the implausilble reaslisations. This should make us somewhat wary of using this model to extrapolate too far beyond the observable range of data

g = sns.relplot(data=summary_df, x="Income", y="estimate", hue="Edu")

g.fig.suptitle("Marginal Predictions of the Outcome Variable \n conditional on counterfactual values for Edu and Income", y=1.05);

The Spline Component Contributions

Finally, we’ll pull out the component contributions of each variable and see how they combine additively. This will also serve as a kind of posterior predictive check for each country as we can show the degree which posterior draws from each component sum to achieve a plausible mirror of the observed data.

Bincome = spline_model.response_component.design.common['bs(Income, degree=3, knots=knots_income)']

income_coefs = az.extract(spline_idata['posterior']['bs(Income, degree=3, knots=knots_income)'])['bs(Income, degree=3, knots=knots_income)']

Bedu = spline_model.response_component.design.common['bs(Edu, degree=3, knots=knots_edu)']

edu_coefs = az.extract(spline_idata['posterior']['bs(Edu, degree=3, knots=knots_edu)'])['bs(Edu, degree=3, knots=knots_edu)']


Bhealth = spline_model.response_component.design.common['bs(Health, degree=3, knots=knots_health)']

health_coefs = az.extract(spline_idata['posterior']['bs(Health, degree=3, knots=knots_health)'])['bs(Health, degree=3, knots=knots_health)']

income = np.dot(Bincome, income_coefs).T 
edu = np.dot(Bedu, edu_coefs).T
health = np.dot(Bhealth, health_coefs).T

intercept = az.extract(spline_idata['posterior']['Intercept'])['Intercept'].values

fig, ax = plt.subplots(figsize=(10, 7))
for i in range(100):
    if i == 1:
        ax.plot(income[i], label='Income Component', color='red')
        ax.plot(edu[i], label='Edu Component', color='blue')
        ax.plot(health[i], label='Health Component', color='darkgreen')
        ax.plot(intercept[i] + income[i] + edu[i] + health[i], label='Combined Components', color='purple')
    else: 
        ax.plot(income[i], alpha=0.1, color='red')
        ax.plot(edu[i], alpha=0.1, color='blue')
        ax.plot(health[i], alpha=0.1, color='darkgreen')
        ax.plot(intercept[i] + income[i] + edu[i] + health[i], color='purple', alpha=0.3)

ax.scatter(range(len(spline_idata['observed_data']['Overall'])), spline_idata['observed_data']['Overall'], label='Observed', color='grey', s=56, ec='black')
ax.set_title("Additive Spline Components", fontsize=20)
ax.legend();
ax.set_xticklabels(pisa_df.dropna(axis=0).reset_index()['Country']);

We can see here how the combined components borrow the structure of the outcome variable primarily from the income variable component. The health measures have closer to zero additive contribution while the uncertainty in the educational component varies wildly. But the blue educational component here is used primarily as a scaling contribution which adds to the level of the outcome variable rather than distorting the shape.

Conclusion

We’ve seen the application of splines as univariate smoothers to approximate wiggly curves of arbitrary shape. We’ve also tried gaussian process approximations of the same univariate functions. The suggested flexibility of both methods is a strength, but we need to be careful where splines have a tendency to over-fit to individual curves. As such we have tried to show that we can incorporate spline basis modelling in a hierarchical bayesian model and recover more compelling posterior predictive checks and additionally derive predictiions from the mixed variant of the hierarhical model which helps us understand the implications of the data for generic forecasts of insurance loss curves. Finally we showed how splines can be used additively to model non-linear functions of multiple covariates. These are a powerful tool to interrogate complex non-linear relationships, but they offer interpolation of functions within a well understood domain. Applealing to these models for extrapolation needs to be done carefully.